Problem: Simplify; express your answer in exponential form. Assume $x\neq 0, k\neq 0$. $\dfrac{{(x^{3})^{2}}}{{(x^{-3}k^{-3})^{-3}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${x^{3}}$ to the exponent ${2}$ . Now ${3 \times 2 = 6}$ , so ${(x^{3})^{2} = x^{6}}$ In the denominator, we can use the distributive property of exponents. ${(x^{-3}k^{-3})^{-3} = (x^{-3})^{-3}(k^{-3})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(x^{3})^{2}}}{{(x^{-3}k^{-3})^{-3}}} = \dfrac{{x^{6}}}{{x^{9}k^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{6}}}{{x^{9}k^{9}}} = \dfrac{{x^{6}}}{{x^{9}}} \cdot \dfrac{{1}}{{k^{9}}} = x^{{6} - {9}} \cdot k^{- {9}} = x^{-3}k^{-9}$.